# (algebra) Proving subspaces- functions

• natalie:)
Oh I understand we can just name two functions since U represents the set of functions...well (f+g)(a) would = (f+g)(b) and would be equivalent to f(a)+g(a)=f(b)+g(b)? I'm not really sure if I'm right about that.(f+g)(a) = f(a) + g(a) = f(b) + g(b) = (f + g)(b)Can you give reasons for each of the = signs above?(f+g)(a) = f(a) + g(a) because that is how the sum of two functions is defined.f(a) + g(a) = f(b) + g(b) because we
natalie:)

## Homework Statement

Is U={f E F($$\left|a,b\right|$$) f(a)=f(b)} a subspace of F($$\left| a,b \right|$$) where F($$\left| a,b \right|$$) is the vector space of real valued functions defined on the interval [a,b]?

## Homework Equations

I know in order for something to be a subspace there are three conditions:
- existence of the zero vector $$\Theta$$
- closed under scalar multiplication

## The Attempt at a Solution

I tried to determine some boundaries for the function (not 1 to 1, all values will be positive, a and b can be any real number). I'm not really sure how to approach it ... i thought maybe if f(a)=0 then f(b) will also =0. But i don't really know how to prove that or the other two conditions.

I would really appreciate someone explaining the steps of how to solve this. Thanks!

Last edited:
natalie:) said:

## Homework Statement

Is U={f E F($$\left|a,b\right|$$) f(a)=f(b)} a subspace of F($$\left| a,b \right|$$) where F($$\left| a,b \right|$$) is the vector space of real valued functions defined on the interval [a,b]?

## Homework Equations

I know in order for something to be a subspace there are three conditions:
- existence of the zero vector $$\Theta$$
- closed under scalar multiplication

## The Attempt at a Solution

I tried to determine some boundaries for the function (not 1 to 1, all values will be positive, a and b can be any real number). I'm not really sure how to approach it ... i thought maybe if f(a)=0 then f(b) will also =0.
Don't put any restrictions such as these on your functions.
natalie:) said:
But i don't really know how to prove that or the other two conditions.

I would really appreciate someone explaining the steps of how to solve this. Thanks!

Clearly U contains the zero function.

Now, let f and g be any arbitrary functions in U, and c any scalar. What can you say about (f + g)(a) and (f + g)(b)?

What can you say about cf(a) and cf(b)?

Oh I understand we can just name two functions since U represents the set of functions...
well (f+g)(a) would = (f+g)(b) and would be equivalent to f(a)+g(a)=f(b)+g(b)? I'm not really sure if I'm right about that.

then c*f(a)=c*f(b)..

I understand the point of what's happening but not the logic behind what's actually happening, if that makes any sense...

natalie:) said:
Oh I understand we can just name two functions since U represents the set of functions...
well (f+g)(a) would = (f+g)(b) and would be equivalent to f(a)+g(a)=f(b)+g(b)? I'm not really sure if I'm right about that.
(f+g)(a) = f(a) + g(a) = f(b) + g(b) = (f + g)(b)
Can you give reasons for each of the = signs above?
What can you conclude about the function f + g?
natalie:) said:
then c*f(a)=c*f(b)..
Or better, (cf)(a) = c*f(a) = c*f(b) = (cf)(b)
Can you give reasons for each = above? What can you conclude about the function cf?
natalie:) said:
I understand the point of what's happening but not the logic behind what's actually happening, if that makes any sense...

Ok well...
(f+g)(a)=f(a)+g(a) i have no idea i want to say distributive property?
f(a)+g(a)= f(b)+g(b) because of the limitations in the question right? if they both have to come to the same solution.. f(a)=f(b) then g(a)=g(b). For some reason all i can think about is extreme value theorem which has nothing to do with this.

The function f+g is... no idea.

For scalar multiplication it makes sense that if you multiply any scalar by the function its the same as multiplying it inside the function c*f(a)=(cf)(a), and again I'm not really sure but think it has something to do with the f(a)=f(b)...

Does the absolute value change anything in solving or does it just imply the functions x-values stay above the axis?

natalie:) said:
Ok well...
(f+g)(a)=f(a)+g(a) i have no idea i want to say distributive property?
No, this is the sum of two functions is defined. The distributive property shows how multiplication distributes over a sum of numbers. There is no multiplication in (f + g)(a).
natalie:) said:
f(a)+g(a)= f(b)+g(b) because of the limitations in the question right?
Sort of. We're assuming that f and g are in U, so f(a) = f(b) and g(a) = g(b). Adding g(a) to f(a) is the same as adding g(b) to f(b).
natalie:) said:
if they both have to come to the same solution.. f(a)=f(b) then g(a)=g(b).
No, one doesn't follow from the other. They both follow from our assumption that f and g are in U.
natalie:) said:
For some reason all i can think about is extreme value theorem which has nothing to do with this.

The function f+g is... no idea.
What you should conclude is that when f and g are in U, then their sum, f + g, will also be in U. Another way to say this is that U is closed under addition.
natalie:) said:
For scalar multiplication it makes sense that if you multiply any scalar by the function its the same as multiplying it inside the function c*f(a)=(cf)(a), and again I'm not really sure but think it has something to do with the f(a)=f(b)...
There is nothing going on "inside the function." c*f(a) = (cf)(a) by the definition of scalar multiples of a function.

Given two functions f and g, you can create four new functions: f + g, f - g, f*g, and f/g. Besides these four using arithmetic operations, there is also the composition f o g, where (f o g)(x) = f(g(x)). And there's the other order, g o f.

natalie:) said:
Does the absolute value change anything in solving or does it just imply the functions x-values stay above the axis?
What absolute value?

Wow I was really not understanding but now really am!
Oh and i was confused at the notation i thought the square brackets were abs value bars but they're brackets

Thank you so much for your help!

natalie:) said:
Wow I was really not understanding but now really am!
Cool!
natalie:) said:
Oh and i was confused at the notation i thought the square brackets were abs value bars but they're brackets

Thank you so much for your help!
You're welcome. I enjoy doing it.

## 1. What is a subspace in algebra?

A subspace in algebra is a subset of a vector space that satisfies all the properties of a vector space. In other words, it is a collection of vectors that can be added together and multiplied by scalars to form new vectors within the same vector space.

## 2. How do you prove that a set of functions is a subspace?

To prove that a set of functions is a subspace, you need to show that it satisfies the three properties of a vector space: closure under addition, closure under scalar multiplication, and contains the zero vector. This can be done by showing that the sum of any two functions in the set is also in the set, and that multiplying any function by a scalar also results in a function within the set.

## 3. Can a set of functions be a subspace if it contains a function that is not continuous?

No, a set of functions cannot be a subspace if it contains a function that is not continuous. Continuity is a necessary property for a function to be a vector in a vector space. Therefore, all functions within a subspace must also be continuous.

## 4. Is the zero function always included in a subspace of functions?

Yes, the zero function (f(x) = 0) is always included in a subspace of functions. This is because the zero function satisfies the properties of a vector space, and is necessary for closure under addition and scalar multiplication.

## 5. Can a set of functions be a subspace if it contains only one function?

Yes, a set of functions can still be a subspace if it contains only one function. As long as that function satisfies the three properties of a vector space, it can be considered a subspace. However, this is a trivial case as any set with only one element can be considered a subspace.

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